Eigenvectors of A'*A for non-square matrix A

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Urs Hackstein
Urs Hackstein 2021 年 5 月 19 日
回答済み: Jaynik 2024 年 3 月 1 日
Let A be a non-square matrix. How can we determine the eigenvector associated with the minimum eigenvalue of the matrix A'*A?
In that paper, it is suggested to use "svd"-function, but how exactly?
  1 件のコメント
David Goodmanson
David Goodmanson 2021 年 5 月 19 日
Hi Urs, you can look up the svd on wikipedia and go to 'Relation to eigenvalue decomposition'

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回答 (1 件)

Jaynik
Jaynik 2024 年 3 月 1 日
Hi,
If you have the matrix A, you can directly use the "eig" function to obtain the eigen vector associated with the minimum eigen value. Following is the code to do the same:
B = A'*A;
[V, D] = eig(B);
[min_eigenvalue, index] = min(diag(D)); % The diagonal of D contains the eigenvalues.
min_eigenvector = V(:, index); % The corresponding column in V is the associated eigenvector.
Alternatively, the "svd" function provides the singular values, which are the square roots of the non-negative eigenvalues of A'*A, and the right singular vectors: Following code can be used for the same:
[U, S, V] = svd(A'*A);
[~, minIndex] = min(diag(S)); % The diagonal elements of S are the square roots of eigenvalues.
min_eigenvector = V(:, minIndex);
You can refer the following documentation to read more about these functions:
Hope this helps!

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